statics chap 3

VECTOR MECHANICS FOR ENGINEERS:

STATICS

Seventh Edition

Ferdinand P. BeerE. Russell Johnston, Jr.

Lecture Notes:J. Walt OlerTexas Tech University

CHAPTER

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

Rigid Bodies: Equivalent Systems of Forces


Vector Mechanics for Engineers: Statics

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ContentsIntroduction

External and Internal Forces

Principle of Transmissibility: Equivalent Forces

Vector Products of Two Vectors

Moment of a Force About a Point

Varigon’s Theorem

Rectangular Components of the Moment of a Force

Sample Problem 3.1

Scalar Product of Two Vectors

Scalar Product of Two Vectors: Applications

Mixed Triple Product of Three Vectors

Moment of a Force About a Given Axis

Sample Problem 3.5

Moment of a Couple

Addition of Couples

Couples Can Be Represented By Vectors

Resolution of a Force Into a Force at O and a Couple

Sample Problem 3.6

System of Forces: Reduction to a Force and a Couple

Further Reduction of a System of Forces

Sample Problem 3.8

Sample Problem 3.10



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Introduction• Treatment of a body as a single particle is not always possible. In

general, the size of the body and the specific points of application of the forces must be considered.

• Most bodies in elementary mechanics are assumed to be rigid, i.e., the actual deformations are small and do not affect the conditions of equilibrium or motion of the body.

• Current chapter describes the effect of forces exerted on a rigid body and how to replace a given system of forces with a simpler equivalent system.

• moment of a force about a point• moment of a force about an axis• moment due to a couple

• Any system of forces acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple.



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External and Internal Forces

• Forces acting on rigid bodies are divided into two groups:

- External forces- Internal forces

• External forces are shown in a free-body diagram.

• If unopposed, each external force can impart a motion of translation or rotation, or both.



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Principle of Transmissibility: Equivalent Forces• Principle of Transmissibility -

Conditions of equilibrium or motion are not affected by transmitting a force along its line of action.NOTE: F and F’ are equivalent forces.

• Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck.

• Principle of transmissibility may not always apply in determining internal forces and deformations.



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Vector Product of Two Vectors• Concept of the moment of a force about a point is

more easily understood through applications of the vector product or cross product.

• Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions:1. Line of action of V is perpendicular to plane

containing P and Q.2. Magnitude of V is3. Direction of V is obtained from the right-hand

rule.

θsinQPV =

• Vector products:- are not commutative,- are distributive,- are not associative,

( )QPPQ ×−=×( ) 2121 QPQPQQP ×+×=+×

( ) ( )SQPSQP ××≠××



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Vector Products: Rectangular Components• Vector products of Cartesian unit vectors,

00

0

=×=×−=×−=×=×=×

=×−=×=×

kkikjjkiijkjjkji

jikkijii

rrrrvrrr

rrrrrrrr

rrrrrrrr

• Vector products in terms of rectangular coordinates

( ) ( )kQjQiQkPjPiPV zyxzyxrrrrrrr

++×++=

( ) ( )( )kQPQP

jQPQPiQPQP

xyyx

zxxzyzzyr

rr

−+

−+−=

zyx

zyxQQQPPPkjirrr

=



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Moment of a Force About a Point• A force vector is defined by its magnitude and

direction. Its effect on the rigid body also depends on it point of application.

• The moment of F about O is defined asFrMO ×=

• The moment vector MO is perpendicular to the plane containing O and the force F.

• Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along MO.

The sense of the moment may be determined by the right-hand rule.

FdrFMO == θsin

• Any force F’ that has the same magnitude and direction as F, is equivalent if it also has the same line of action and therefore, produces the same moment.



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Moment of a Force About a Point

• Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained in the plane of the structure.

• The plane of the structure contains the point O and the force F. MO, the moment of the force about O is perpendicular to the plane.

• If the force tends to rotate the structure clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive.

• If the force tends to rotate the structure counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative.



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Varignon’s Theorem

• The moment about a give point O of the resultant of several concurrent forces is equal to the sum of the moments of the various moments about the same point O.

( ) Lrrrr

Lrrr +×+×=++× 2121 FrFrFFr

• Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F.



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( ) ( ) ( )kyFxFjxFzFizFyF

FFFzyxkji

kMjMiMM

xyzxyz

zyx

zyxO

rrr

rrr

rrrr

−+−+−=

=

++=

The moment of F about O,

kFjFiFFkzjyixrFrM

zyx

O rrrr

rrrrrrr

++=++=×= ,



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The moment of F about B,

FrM BAB

rrr×= /

( ) ( ) ( )kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

rrrr

rrr

rrr

++=

−+−+−=

−=/

( ) ( ) ( )zyx

BABABAB

FFFzzyyxx

kjiM −−−=

rrr

r



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Rectangular Components of the Moment of a ForceFor two-dimensional structures,

( )

zy

ZO

zyO

yFxFMM

kyFxFM

−==

−=rr

( ) ( )[ ]

( ) ( ) zBAyBA

ZO

zBAyBAO

FyyFxxMM

kFyyFxxM

−−−==

−−−=rr



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Sample Problem 3.1

A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O.

Determine:

a) moment about O,

b) horizontal force at A which creates the same moment,

c) smallest force at A which produces the same moment,

d) location for a 240-lb vertical force to produce the same moment,

e) whether any of the forces from b, c, and d is equivalent to the original force.



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Sample Problem 3.1

a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper.

( )( )( )in. 12lb 100

in. 1260cosin.24=

=°==

O

O

Md

FdM

in lb 1200 ⋅=OM



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Sample Problem 3.1

c) Horizontal force at A that produces the same moment,

( )

( )

in.8.20in. lb 1200

in. 8.20in. lb 1200

in.8.2060sinin.24

⋅=

=⋅=

=°=

F

FFdM

d

O

lb 7.57=F



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Sample Problem 3.1c) The smallest force A to produce the same moment

occurs when the perpendicular distance is a maximum or when F is perpendicular to OA.

( )

in.42in. lb 1200

in. 42in. lb 1200⋅

=

=⋅=

F

FFdMO

lb50=F



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Sample Problem 3.1

d) To determine the point of application of a 240 lb force to produce the same moment,

( )

in. 5cos60

in. 5lb 402

in. lb 1200lb 240in. lb 1200

=°

=⋅

=

=⋅=

OB

d

dFdMO

in.10=OB



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Sample Problem 3.1

e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 lb force.



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Sample Problem 3.4

SOLUTION:

The moment MA of the force F exerted by the wire is obtained by evaluating the vector product,

FrM ACA

rrr×=

The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C.



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Sample Problem 3.4SOLUTION:

1289612008.003.0

−−=

kjiM A

rrr

r

( ) ( ) ( )kjiM A

rrrvmN 8.82mN 8.82mN 68.7 ⋅+⋅+⋅−=

( ) ( ) jirrr ACAC

rrrrr m 08.0m 3.0 +=−=

FrM ACA

rrr×=

( )

( ) ( ) ( ) ( )

( ) ( ) ( )kji

kji

rr

FFDC

DC

rrr

rrr

rrr

N 128N 69N 120m 5.0

m 32.0m 0.24m 3.0N 200

N 200

−+−=

−+−=

== λ



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Scalar Product of Two Vectors• The scalar product or dot product between

two vectors P and Q is defined as( )resultscalarcosθPQQP =•

rr

• Scalar products:- are commutative,- are distributive,- are not associative,

PQQPrrrr

•=•( ) 2121 QPQPQQP

rrrrrrr•+•=+•

( ) undefined =•• SQPrrr

• Scalar products with Cartesian unit components,

000111 =•=•=•=•=•=• ikkjjikkjjiirrrvrrrrrrrr

( ) ( )kQjQiQkPjPiPQP zyxzyx

rrrrrrrr++•++=•

2222 PPPPPP

QPQPQPQP

zyx

zzyyxx

=++=•

++=•rr

rr



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Scalar Product of Two Vectors: Applications• Angle between two vectors:

PQQPQPQP

QPQPQPPQQP

zzyyxx

zzyyxx

++=

++==•

θ

θ

cos

cosrr

• Projection of a vector on a given axis:

OL

OL

PPQ

QPPQQP

OLPPP

==•

=•

==

θ

θ

θ

cos

cos

alongofprojection cos

rr

rr

zzyyxx

OL

PPPPP

θθθλ

coscoscos ++=•=rr

• For an axis defined by a unit vector:



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Mixed Triple Product of Three Vectors• Mixed triple product of three vectors,

( ) resultscalar =×• QPSrrr

• The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign,

( ) ( ) ( )( ) ( ) ( )SPQQSPPQS

PSQSQPQPSrrrrrrrr

rrrrrrrrr

×•−=×•−=×•−=

×•=×•=×•

( ) ( ) ( )( )

zyx

zyx

zyx

xyzxyz

zxxzyyzzyx

QQQPPPSSS

QPQPS

QPQPSQPQPSQPS

=

−+

−+−=×•rrr

• Evaluating the mixed triple product,



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Moment of a Force About a Given Axis• Moment MO of a force F applied at the point A

about a point O,FrM O

rrr×=

• Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis,

( )FrMM OOL

rrrrr×•=•= λλ

• Moments of F about the coordinate axes,

xyz

zxy

yzx

yFxFM

xFzFM

zFyFM

−=

−=

−=



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Moment of a Force About a Given Axis

• Moment of a force about an arbitrary axis,

( )BABA

BA

BBL

rrr

Fr

MM

rrr

rrr

rr

−=

×•=

•=

λ

λ

• The result is independent of the point Balong the given axis.



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Sample Problem 3.5

A cube is acted on by a force P as shown. Determine the moment of P

a) about Ab) about the edge AB andc) about the diagonal AG of the cube.d) Determine the perpendicular distance

between AG and FC.



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Sample Problem 3.5• Moment of P about A,

( )( ) ( )( ) ( )jiPjiaM

jiPjiPP

jiajaiar

PrM

A

AF

AFA

rrrrr

rrrrr

rrrrr

rrr

+×−=

+=+=

−=−=

×=

2

222

( )( )kjiaPM A

rrrr++= 2

• Moment of P about AB,

( )( )kjiaPi

MiM AABrrrr

rr

++•=

•=

2

2aPM AB =



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Sample Problem 3.5• Moment of P about the diagonal AG,

( )

( )

( ) ( )

( )1116

2312

31

3

−−=

++•−−=

++=

−−=−−

==

•=

aP

kjiaPkjiM

kjiaPM

kjia

kajaiarr

MM

AG

A

GA

GA

AAG

rrrrrr

rrrr

rrrrrrr

r

rr

λ

λ

6aPM AG −=



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Sample Problem 3.5• Perpendicular distance between AG and FC,

( ) ( ) ( )

0

11063

12

=

+−=−−•−=•PkjikjPP

rrrrrrrλ

Therefore, P is perpendicular to AG.

PdaPM AG ==6

6ad =



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Moment of a Couple• Two forces F and -F having the same magnitude,

parallel lines of action, and opposite sense are said to form a couple.

• Moment of the couple,

( )( )

FdrFMFr

FrrFrFrM

BA

BA

==×=

×−=

−×+×=

θsin

rr

rrr

rrrrr

• The moment vector of the couple is independent of the choice of the origin of the coordinate axes, i.e., it is a free vector that can be applied at any point with the same effect.



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Moment of a Couple

Two couples will have equal moments if

• 2211 dFdF =

• the two couples lie in parallel planes, and

• the two couples have the same sense or the tendency to cause rotation in the same direction.



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Addition of Couples• Consider two intersecting planes P1 and

P2 with each containing a couple

222

111

planein planein PFrMPFrM

rrr

rrr

×=

×=

• Resultants of the vectors also form a couple

( )21 FFrRrMrrrrrr

+×=×=

• By Varigon’s theorem

21

21

MMFrFrM

rr

rrrrr

+=

×+×=

• Sum of two couples is also a couple that is equal to the vector sum of the two couples



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Couples Can Be Represented by Vectors

• A couple can be represented by a vector with magnitude and direction equal to the moment of the couple.

• Couple vectors obey the law of addition of vectors.

• Couple vectors are free vectors, i.e., the point of application is not significant.

• Couple vectors may be resolved into component vectors.



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• Force vector F can not be simply moved to O without modifying its action on the body.

• Attaching equal and opposite force vectors at O produces no net effect on the body.

• The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couple system.



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• Moving F from A to a different point O’ requires the addition of a different couple vector MO’

FrM O

rrr×′='

• The moments of F about O and O’ are related,( )

FsM

FsFrFsrFrM

O

Orrr

rrrrrrrrrr

×+=

×+×=×+=×= ''

• Moving the force-couple system from O to O’ requires the addition of the moment of the force at O about O’.



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• Attach equal and opposite 20 lb forces in the +x direction at A, thereby producing 3 couples for which the moment components are easily computed.

• Alternatively, compute the sum of the moments of the four forces about an arbitrary single point. The point D is a good choice as only two of the forces will produce non-zero moment contributions..

Determine the components of the single couple equivalent to the couples shown.



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Sample Problem 3.6

• Attach equal and opposite 20 lb forces in the +x direction at A

• The three couples may be represented by three couple vectors,

( )( )( )( )( )( ) in.lb 180in. 9lb 20

in.lb240in. 12lb 20in.lb540in.18lb30

⋅+=+=

⋅+=+=⋅−=−=

z

y

x

M

MM

( ) ( )( )k

jiMr

rrr

in.lb 180

in.lb240in.lb 540

⋅+

⋅+⋅−=



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Sample Problem 3.6• Alternatively, compute the sum of the

moments of the four forces about D.

• Only the forces at C and E contribute to the moment about D.

( ) ( )( ) ( )[ ] ( )ikj

kjMM Drrr

rrrr

lb 20in. 12in. 9

lb 30in. 18

−×−+

−×==

( ) ( )( )k

jiMr

rrr

in.lb 180

in.lb240in.lb 540

⋅+

⋅+⋅−=



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System of Forces: Reduction to a Force and Couple

• A system of forces may be replaced by a collection of force-couple systems acting a given point O

• The force and couple vectors may be combined into a resultant force vector and a resultant couple vector,

( )∑∑ ×== FrMFR RO

rrrrr

• The force-couple system at O may be moved to O’with the addition of the moment of R about O’ ,

RsMM RO

RO

rrrr×+='

• Two systems of forces are equivalent if they can be reduced to the same force-couple system.



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Further Reduction of a System of Forces• If the resultant force and couple at O are mutually

perpendicular, they can be replaced by a single force acting along a new line of action.

• The resultant force-couple system for a system of forces will be mutually perpendicular if:1) the forces are concurrent, 2) the forces are coplanar, or 3) the forces are parallel.



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Further Reduction of a System of Forces

• System of coplanar forces is reduced to a force-couple system that is mutually perpendicular.

ROMRrr

and

• System can be reduced to a single force by moving the line of action of until its moment about O becomes R

OMrRr

• In terms of rectangular coordinates,ROxy MyRxR =−



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Sample Problem 3.8

For the beam, reduce the system of forces shown to (a) an equivalent force-couple system at A, (b) an equivalent force couple system at B, and (c) a single force or resultant.

Note: Since the support reactions are not included, the given system will not maintain the beam in equilibrium.

SOLUTION:

a) Compute the resultant force for the forces shown and the resultant couple for the moments of the forces about A.

b) Find an equivalent force-couple system at B based on the force-couple system at A.

c) Determine the point of application for the resultant force such that its moment about A is equal to the resultant couple at A.



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a) Compute the resultant force and the resultant couple at A.

( ) ( ) ( ) ( ) jjjjFR

rrrr

rr

N 250N 100N 600N 150 −+−=

= ∑

( ) jRrr

N600−=

( )( ) ( ) ( ) ( )

( ) ( )jijiji

FrM RA

rr

rrrr

rrr

2508.41008.26006.1

−×+

×+−×=

×= ∑

( )kM RA

rrmN 1880 ⋅−=



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Sample Problem 3.8b) Find an equivalent force-couple system at B

based on the force-couple system at A.

The force is unchanged by the movement of the force-couple system from A to B.

( ) jRrr

N 600−=

The couple at B is equal to the moment about Bof the force-couple system found at A.

( ) ( ) ( )( ) ( )kk

jik

RrMM ABRA

RB

rr

rrr

rrrr

mN 2880mN 1880

N 600m 8.4mN 1880

⋅+⋅−=

−×−+⋅−=

×+=

( )kM RB

rrmN 1000 ⋅+=



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Sample Problem 3.10

SOLUTION:

• Determine the relative position vectors for the points of application of the cable forces with respect to A.

• Resolve the forces into rectangular components.

• Compute the equivalent force,

∑= FRrr

Three cables are attached to the bracket as shown. Replace the forces with an equivalent force-couple system at A.

• Compute the equivalent couple,

( )∑ ×= FrM RA

rrv



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Sample Problem 3.10

SOLUTION:

• Determine the relative position vectors with respect to A.

( )( )( )m 100.0100.0

m 050.0075.0

m 050.0075.0

jir

kir

kir

AD

AC

AB

rrr

rrr

rrr

−=

−=

+=

• Resolve the forces into rectangular components.

( )

( )N 200600300

289.0857.0429.0

1755015075

N 700

kjiF

kji

kjirr

F

B

BE

BE

B

rrrr

rrr

rrrrr

rr

+−=

+−=

+−==

=

λ

λ

( )( )( )N 1039600

30cos60cosN 1200ji

jiFDrr

rrr

+=

+=

( )( )( )N 707707

45cos45cosN 1000ji

jiFCrr

rrr

−=

−=



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Sample Problem 3.10• Compute the equivalent force,

( )( )( )k

ji

FR

r

r

r

rr

707200

1039600600707300

−+

+−+

++=

= ∑

( )N 5074391607 kjiRrrrr

−+=

• Compute the equivalent couple,

( )

kkji

Fr

jkji

Fr

kikji

Fr

FrM

DAD

cAC

BAB

RA

r

rrr

rr

r

rrr

rr

rr

rrr

rr

rrv

9.163010396000100.0100.0

68.177070707050.00075.0

4530200600300050.00075.0

=−=×

=−−=×

−=−

=×

×= ∑

kjiM RA

rrrr9.11868.1730 ++=

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